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# 数学代写|线性代数代写Linear algebra代考|Finite dimensional vector space

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## 数学代写|线性代数代写Linear algebra代考|Finite dimensional vector space

What does the term finite dimensional vector space mean?
It is a vector space $V$ which has a finite number of vectors in its basis.
Definition (3.18). In general, if a finite number of vectors form a basis for a vector space $V$ then we say $V$ is finite dimensional. Otherwise, the vector space $V$ is known as infinite dimensional.
If the vector space $V$ consists only of the zero vector then it is also finite dimensional.
Can you think of any finite dimensional vector spaces?
The Euclidean spaces $-\mathbb{R}^2, \mathbb{R}^3, \mathbb{R}^4, \ldots, \mathbb{R}^n$.

Are there any other examples of finite dimensional vector spaces?
The set $P_2$ of polynomials of degree 2 or less for example, or the set of all 2 by 2 matrices $M_{22}$. (These were covered in the previous section.)

Definition (3.19). In general, if $n$ vectors $\left{\mathbf{v}1, \mathbf{v}_2, \mathbf{v}_3, \ldots, \mathbf{v}_n\right}$ form a basis for a vector space $V$ then we say that $V$ is $n$-dimensional. What is $\operatorname{dim}\left(M{22}\right)$ equal to?
The standard basis for $M_{22}$ (matrices of size 2 by 2) from the Exercises 3.3 question 5 is
$$\mathbf{A}=\left(\begin{array}{ll} 1 & 0 \ 0 & 0 \end{array}\right), \mathbf{B}=\left(\begin{array}{ll} 0 & 1 \ 0 & 0 \end{array}\right), \mathbf{C}=\left(\begin{array}{ll} 0 & 0 \ 1 & 0 \end{array}\right) \text { and } \mathbf{D}=\left(\begin{array}{ll} 0 & 0 \ 0 & 1 \end{array}\right)$$
Therefore $\operatorname{dim}\left(M_{22}\right)=4$ because we have four matrices in ${\mathbf{A}, \mathbf{B}, \mathbf{C}, \mathbf{D}}$ which form a basis for $M_{22}$.
What is $\operatorname{dim}\left(P_2\right)$ equal to?
Remember that the standard basis for $P_2$ (the set of all polynomials of degree 2 or less) is the set $\left{1, t, t^2\right}$, which means $\operatorname{dim}\left(P_2\right)=3$ since the basis consists of three vectors.
Table 3.1 shows some vector spaces and their dimensions.

## 数学代写|线性代数代写Linear algebra代考|Properties of finite dimensional vector spaces

In this section we show some important properties of bases and dimension. This is a demanding section because the proofs of propositions are lengthy.
Lemma (3.21). Let $V$ be a finite $n$-dimensional vector space. We have the following:
(a) Let $\left{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \ldots, \mathbf{v}_n\right}$ be a set of linearly independent vectors. Then $\left{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_m\right}$ where $m>n(m$ is greater than $n)$ is linearly dependent.
(b) If the $n$ vectors $\left{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3, \ldots, \mathbf{u}_n\right}$ span $V$ then $\left{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3, \ldots, \mathbf{u}_m\right}$ where $m<n$ does not $\operatorname{span} V$.
What does part (a) mean?
In $n$-dimensional vector space, if you add additional vectors to $n$ linearly independent vectors then the set becomes linearly dependent.
How do we prove this result?
Proof of $(a)$.
The number of basis vectors in $V$ is $n$. Suppose that $\left{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_m\right}$ are linearly independent. Then we must have $m \leq n$ ( $m$ is less than or equal to $n)$.
Why?
Because by question 14 of Exercises 3.3 we know that the number of vectors in a linearly independent set must be less than or equal to $n, m \leq n$ (the number of basis vectors).

However, we are given that $m>n$ ( $m$ is greater than $n)$, therefore our supposition that $\left{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_m\right}$ is a linearly independent set of vectors must be wrong, so this set is linearly dependent.

## 数学代写|线性代数代写Linear algebra代考|Finite dimensional vector space

$$\mathbf{A}=\left(\begin{array}{ll} 1 & 0 \ 0 & 0 \end{array}\right), \mathbf{B}=\left(\begin{array}{ll} 0 & 1 \ 0 & 0 \end{array}\right), \mathbf{C}=\left(\begin{array}{ll} 0 & 0 \ 1 & 0 \end{array}\right) \text { and } \mathbf{D}=\left(\begin{array}{ll} 0 & 0 \ 0 & 1 \end{array}\right)$$

$\operatorname{dim}\left(P_2\right)$等于多少?

## 数学代写|线性代数代写Linear algebra代考|Properties of finite dimensional vector spaces

(a)设$\left{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \ldots, \mathbf{v}_n\right}$为线性无关向量的集合。然后$\left{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_m\right}$，其中$m>n(m$大于$n)$是线性相关的。
(b)如果$n$向量$\left{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3, \ldots, \mathbf{u}_n\right}$跨越$V$，则$\left{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3, \ldots, \mathbf{u}_m\right}$，其中$m<n$不$\operatorname{span} V$。
(a)部分是什么意思?

$V$中基向量的个数为$n$。假设$\left{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_m\right}$是线性无关的。那么我们就得到$m \leq n$ ($m$小于等于$n)$)

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